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The null expansion, null shear and null twist of a congruence tangent to the generators of a Killing horizon all vanish there. If $k$ is its Killing generator and $\ell =fk$ is an affine generator, the screen-space projector kills every term proportional to $k$. The symmetric part of the projected derivative of $\ell$ therefore vanishes by the Killing equation. The antisymmetric part vanishes by Frobenius theorem, since $\ell$ is hypersurface-orthogonal. These conclusions extend through a regular bifurcation surface by continuity.

In the Ingoing Eddington-Finkelstein coordinates metric $d{s}^2=-f(r)d{v}^2+2dvdr+r^{2}d{\Omega }^2$, where $f=1-2M/r$, the tangent $U=-E{\partial }_r$ is affine when $E>0$ is constant along each ray. The displayed radial partner obeys $N^2=0$ and $U\cdot N=-1$. Since ${\Gamma }^v{}_{rv}={\Gamma }^a{}_{rr}=0$ and ${\Gamma }^r{}_{rv}=-f^{\prime }/2$, its derivative ${\partial }_r{N}^r=f^{\prime }/(2E)$ cancels the connection term, giving ${\nabla }_{U}N=0$. Its screen is the angular two-plane, with optical tensor ${\hat{B}}_{AB}=-Er{\gamma }_{AB}$, null expansion $-2E/r$, and zero null shear and null twist.

Hawking's area theorem states that the area of later cross-sections of a classical future event horizon cannot be smaller than that of earlier cross-sections, including the sum over initially disconnected components. Its hypotheses include the null energy condition, the Einstein equations, and the global regularity/predictability conditions that exclude future naked pathologies of the horizon generators. A convenient sufficient version assumes those generators are future complete. Without the energy and global hypotheses the assertion is not unconditional.

Let $k^a$ be an affinely parametrized null generator, with expansion $\theta$. Because the horizon is a null hypersurface, the generator congruence has zero null twist. The Null Raychaudhuri equation in four dimensions givesThe null shear square is nonnegative and the null energy condition makes $R_{ab}{k}^a{k}^b=8\pi G{T}_{ab}{k}^a{k}^b\ge 0$; a cosmological term has zero null contraction. Suppose $\theta ({\lambda }_0)<0$. Integration of this inequality forces $\theta \to -\infty$ within affine distance at most $2/|\theta ({\lambda }_0)|$. This produces a focal point to the initial horizon cross-section. Beyond it the null geodesic cannot remain on an achronal boundary, whereas an event horizon is precisely such a boundary. Future completeness, or the corresponding standard predictability condition ensuring continuation over this interval, gives the contradiction. HenceThis proves the local area increase along every smooth generator bundle. Generators can enter the horizon at past crease points, but cannot leave through future endpoints on a regular horizon under these hypotheses. New generators add area, rather than subtracting it, so the result extends across the nonsmooth merger set and to whole horizon cross-sections. This is the future-complete horizon focusing proof of the area theorem.

Choose an early cross-section with the two separate components of areas $A_1,A_2$, and a later cross-section of the settled merged hole. Following the old generators forward maps their area elements injectively into the later horizon; a future crossing/focal point would violate the same achronality argument. Each element has nondecreased area. Any additional generators contribute nonnegative area, givingThis is a comparison of event-horizon cross-sections, not an assertion that set containment alone bounds the areas of arbitrary apparent horizons.

For a Schwarzschild black hole, $r_h=2Gm/c^2$ and $A=4\pi r_h^2=16\pi G^2{m}^2/c^4$. Applying the merger inequality to the two equal initial masses and the final Schwarzschild mass givesFor initially well-separated holes at rest, with no extra incoming energy or external work, energy conservation gives $E_{\mathrm{rad}}=(2m-M)c^2$. The area bound on equal-mass Schwarzschild merger radiation is consequentlyThis is the maximum permitted by the area/energy inequalities, not a demonstration that a dynamical merger attains it. A reversible limiting area change would be needed to saturate the bound. If initial separation, binding energy or external agents change the initial total energy, replace $2m{c}^2$ by that energy when computing escaped radiation. Quantum evaporation does not contradict the theorem: the classical energy-condition hypotheses are not retained unchanged in that setting.

Use Planck units for the following formulas. The black-hole area theorem is a classical statement with hypotheses. Assume the Einstein field equations, the null energy condition and appropriate global future regularity/predictability. For a clean version of the focusing sketch, take the future horizon generators to remain on a regular horizon and to be future complete in affine parameter. The usual strong asymptotic predictability formulation supplies the global causal control needed to exclude the same focusing pathology; an arbitrary spacetime with naked future breakdown is not covered by the theorem.

Let ${\ell }^a$ be an affinely parametrized generator of a smooth portion of the event horizon, and let $\theta =d\log dA/d\lambda$ be its null expansion. The horizon is a null hypersurface, so its generators have zero null twist. In four dimensions the Null Raychaudhuri equation isThe Einstein field equations give $R_{ab}{\ell }^a{\ell }^b=8\pi T_{ab}{\ell }^a{\ell }^b\ge 0$, since the metric terms vanish when contracted with a null vector. The screen-space null shear term is also nonnegative, so ${\theta }^{\prime }\le -{\theta }^2/2$. If ${\theta }_0<0$ at ${\lambda }_0$, integrating this inequality while the congruence remains regular givesIt focuses within affine distance at most $2/|{\theta }_0|$. The vanishing area element is a focal point to an earlier spacelike horizon cut. Past such a point the null geodesic can no longer generate an achronal boundary, but the event horizon is precisely such a boundary. Future completeness and the global hypotheses exclude its escaping this contradiction by ending prematurely at a pathology. Therefore $\theta \ge 0$ everywhere the smooth horizon description applies.

Integrate $d(dA)/d\lambda =\theta dA$ between ordered horizon cuts. Every continuing generator contributes a nondecreasing area element. New generators can enter at past endpoints or merger crease sets, adding area; generators cannot leave a regular future horizon through future endpoints under these hypotheses. Thus, including the total area of disconnected components,This is the future-complete horizon focusing proof of the area theorem. A smooth elementary congruence proof requires the stated regularity; the general theorem also treats the horizon's nonsmooth joining set rather than assuming mergers have a globally smooth horizon.

Hawking radiation changes the situation because the quantum field state in a collapsing geometry is not a classical positive-energy fluid. For a large isolated Schwarzschild black hole,in natural units. Outgoing quantum modes at future null infinity have a nearly thermal occupation, modified by greybody factors; an accompanying negative Killing energy flux near the horizon reduces the hole's mass. On timescales short compared with the evaporation time but long compared with $M$, its geometry is approximately Schwarzschild spacetime with slowly decreasing $M(t)$. It is not an exactly stationary vacuum solution with a parameter changed by hand: the radiation's stress-energy tensor sources the time dependence through the semiclassical Einstein equation. The distant luminosity is carried by the outgoing radiation, while the horizon-area decrease accompanies the inward negative-energy contribution.

As the black hole loses mass, its horizon scale and Bekenstein-Hawking entropy shrink, its temperature rises and its luminosity increases. This is the negative heat capacity of a Schwarzschild black hole. Large astrophysical holes evaporate extremely slowly in isolation; small holes evaporate more rapidly, and increasingly massive particle species become accessible as the temperature rises. An incoming radiation bath or accretion can instead offset the loss, so the pure evaporation law assumes negligible incoming energy. A formal complete evaporation also raises the black hole information paradox; the low-curvature calculation alone does not determine how information or the final state is resolved.

For an elementary luminosity estimate, treat the horizon as emitting photon blackbody radiation. In natural units the Stefan–Boltzmann law has $\sigma ={\pi }^2/60$, givingThe horizon area used here is an estimate of emitting area, not the exact frequency-dependent absorption cross-section. More generally write $P=\alpha /M^2$, where $\alpha >0$ includes the greybody factors and the active particle species. For constant effective $\alpha$, energy balance and direct integration giveHere $t$ is time measured at infinity. The formal evaporation timescale is $M_0^3/(3\alpha )$; with the photon blackbody estimate it is $5120\pi M_0^3$. Restoring constants givesfor that specified estimate. If particle thresholds make $\alpha$ mass dependent, the precise leading law is instead $t={\int }_{M(t)}^{M_0}{m}^{2}dm/\alpha (m)$. The robust result is the inverse-square mass-loss rate for a fixed species regime and the cubic mass scaling of its lifetime. This is the semiclassical cubic mass law for Schwarzschild evaporation. Extrapolating its zero to a definite end state is unjustified once $M$ approaches the Planck mass and curvature requires quantum gravity.

There is no contradiction with the classical area theorem: the renormalized quantum stress-energy tensor can violate the null energy condition used in the focusing inequality. In the evaporation regime,which is permitted when that hypothesis fails. The thermodynamic replacement is the generalized second law, concerning $S_{\mathrm{outside}}+A/(4G\hslash )$ rather than horizon area alone. Outside radiation can carry entropy while the hole's Bekenstein-Hawking entropy decreases.

Choose a parallel-transported orthonormal basis of the screen. In that basis a symmetric trace-free null shear and antisymmetric null twist have formsDirect multiplication givesHenceSince the tidal matrix is symmetric, the antisymmetric part of the Sachs optical equations immediately givesThus an initially null twist-free null geodesic congruence remains null twist-free while its optical description is regular.

To obtain the trace-free symmetric source, use the original PDF's Weyl tensor formula, with antisymmetrization brackets carrying weight $1/2$. The TeX aid has lost indices and brackets. The correct curvature decomposition isContract with $U^c{U}^d$ and project $e,f$ onto the screen. Terms containing $U^2$ or a remaining screen-projected $U$ vanish. The only Ricci term left isThe Weyl term has zero screen trace, by its trace-free property and antisymmetry. Consequently the trace-free symmetric part of the optical equation is the null-shear propagation equationThe two-dimensional matrix identities are essential: in a higher-dimensional screen the null shear square can have a trace-free part and the null shear–null twist anticommutator need not vanish. Choosing the parallel-transported auxiliary null vector makes these tensor transport formulae hold without extra moving-screen terms.

Take the screen trace of the optical matrix of a null congruence equation. The cross-contractions of symmetric null shear with antisymmetric null twist vanish, andThe minus sign in the null twist contribution follows from tracing the square of an antisymmetric matrix. The tidal trace is $P^{ef}{R}_{cedf}{U}^c{U}^d=R_{cd}{U}^c{U}^d$: the additional null terms in $P^{ef}$ vanish by curvature antisymmetry. Thus the Null Raychaudhuri equation isIn particular, null shear focuses the null geodesic congruence while null twist enters with the opposite sign. With vanishing null twist and the null energy condition, the Einstein field equations imply $U\cdot \nabla \theta \le -{\theta }^2/2$. These signs and the factor $1/2$ depend on using the full-trace null expansion in a two-dimensional screen.

First let $B_{ab}={\nabla }_b{U}_a$ without projection. Commute derivatives on this covector and use affine geodesic motion:The last equality relabels the two contracted null-vector indices. Nullness and affine parametrization give $U^a{B}_{ab}=0$ and $B_{ab}{U}^b=0$. Therefore inserting $P$ into the contracted middle index changes no screen-projected quadratic term. Since ${\nabla }_{U}P=0$, screen projection proves the Sachs optical equationsThe optical tidal matrix ${\mathcal{R}}_{ab}$ is symmetric by the Riemann curvature tensor's pair symmetries. Raising one screen index gives the corresponding equation for ${\hat{B}}^a{}_b$.

In terms of the defined optical scalars, the equation is explicitlyThis matrix identity contains the null expansion, null shear and null twist evolution equations as its trace, symmetric trace-free part and antisymmetric part.

Choose an auxiliary null vector $N$ with $U\cdot N=-1$, parallel-transported along each affine ray. The screen-space projector and its induced positive metric areThey annihilate $U,N$, obey $P^2=P$, and have trace two in four spacetime dimensions. Since both $U$ and $N$ are parallel along $U$, ${\nabla }_{U}P=0$. Equivalently, the screen represents the two-dimensional quotient $U^{\perp }/\mathrm{span}(U)$.

Define the optical matrix of a null congruence byFor a transverse deviation, its screen projection $\zeta =PZ$ satisfies ${\nabla }_U{\zeta }^a={\hat{B}}^a{}_b{\zeta }^b$. Decompose the matrix asThese are the null expansion, null shear and null twist. Null expansion gives the fractional change of a small transverse area, $\theta =d\log (dA)/d\lambda$. Null shear is symmetric trace-free distortion; null twist is the antisymmetric rotation part. This null expansion convention uses the full trace, not half the trace. The auxiliary screen fixes representatives, while the transverse optical content is intrinsic to the null geodesic congruence.

Use geometrized units $G=c=1$. The Physical-process first law of black-hole mechanics concerns a small perturbation of a rotating black hole in an initially stationary spacetime by infalling matter, followed by relaxation to another stationary spacetime. For an uncharged black hole, or neutral accretion with charge held fixed, its first-order statement isHere $\kappa >0$ is the background surface gravity, ${\Omega }_H$ is the background horizon angular velocity, and $\delta A$ is the change in event horizon area. The energy $\delta E$ and angular momentum $\delta J$ are the fluxes carried into the event horizon; when these account for the changes of the black hole, they equal $\delta M$ and its change of angular momentum. The result is a linearized law, not an exact identity for a finite violent accretion event. The perturbation must be small enough that focusing does not produce new caustics or invalidate the expansion about the existing generators.

Let $k^a$ be the stationary Killing vector field normalized at infinity, $m^a$ the axial Killing vector field with $2\pi$-periodic orbits, andOn the background Killing horizon, ${\chi }^a$ is its future-directed null generator and satisfies ${\chi }^b{\nabla }_b{\chi }^a=\kappa {\chi }^a$. Choose a parameter $v$ with $\chi ={\partial }_v$. The null expansion in this parameter is $\theta ={\partial }_v\log dA$. The generator is not affinely parametrized; if $\ell ={\partial }_{\lambda }$ is affine, then $\chi =\kappa \lambda \ell$ after an appropriate choice of affine origin, with $\lambda \propto e^{\kappa v}$. Rescaling the Null Raychaudhuri equation givesThe null twist vanishes because the generators are normal to the event horizon. By the Einstein field equations, $R_{ab}{\chi }^a{\chi }^b=8\pi T_{ab}{\chi }^a{\chi }^b$. Background null expansion and null shear vanish on a Killing horizon of the background stationary spacetime. Thus $\theta$ and $\hat{\sigma }$ are first order, their squares are second order, and the linearized equation isThe final stationary spacetime supplies the boundary condition $\theta \to 0$ in the future. Explicitly,This future boundary condition reflects the global definition of the event horizon. For localized infall, $\theta$ also tends to zero in the remote past. Integrating the linearized equation along the generators, and then over a background cross-section with area element $d{A}_0$, givesUsing the background area element introduces only second-order corrections.

With orientations chosen so that future infalling positive energy has positive flux, the conserved Killing energy and axial angular momentum giveConsequently $\delta E-{\Omega }_H\delta J={\int }_{\mathcal{H}}\delta T_{ab}{\chi }^a{\chi }^{b}dvd{A}_0$, proving the displayed Physical-process first law of black-hole mechanics. Restoring $G$ replaces $8\pi$ by $8\pi G$. The null energy condition makes the integral nonnegative and therefore implies area increase in this regime; that condition is needed for the sign conclusion, not for the linearized flux identity itself. Charged infall would require the additional electrostatic work term, and an extremal background needs separate treatment because the proof uses $\kappa >0$.

A null geodesic congruence is a smooth family of null geodesics filling a region, with one curve through each point before caustics form. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^2=0$ and ${\nabla }_{U}U=0$. Choose a null partner $N^a$ with $U\cdot N=-1$, transported parallel to $U$ along each generator. The screen-space projectorprojects onto the positive-definite two-dimensional plane transverse to $U,N$. Define the optical tensor and its decomposition bywhereThese are respectively the null expansion, null shear and null twist, the latter also called rotation. In the Null Raychaudhuri equation, ${\hat{\sigma }}^2={\hat{\sigma }}_{ab}{\hat{\sigma }}^{ab}$ and ${\hat{\omega }}^2={\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$ use the screen metric. The trace convention for $\theta$ is important: it is not half the trace. For an affine null tangent it also equals ${\nabla }_a{U}^a$.

Transport two connecting vectors between neighbouring rays so that they commute with $U$, and project them onto the screen. Their infinitesimal transverse lengths and angles evolve according to the symmetric part of the optical tensor. The trace part gives uniform fractional dilation, while the trace-free null shear stretches one transverse direction and contracts the other, changing a circular beam to an ellipse without changing its area to first order. With zero null twist, there is no antisymmetric rotation of the beam.

If $q_{AB}$ is the metric on an infinitesimal transverse element, then $d{q}_{AB}/d\lambda =2{\hat{B}}_{(AB)}$. Its area $A$ therefore obeysThus positive null expansion means a beam's cross-sectional area increases, and negative expansion means it decreases; null shear records the directional distortion rather than this overall area change.

Choose the affine tangent $k^a$ to a null geodesic congruence, and an auxiliary null vector ${\ell }^a$ with $k\cdot \ell =-1$. The screen-space projectorannihilates $k$ and $\ell$ and restricts to a positive-definite two-dimensional metric in four spacetime dimensions. Neighboring transverse rays have relative displacement governed by the optical tensorIts irreducible decomposition defines the three requested optical quantities:The null expansion $\theta$ is the fractional rate of change of transverse area, $d\log A/d\lambda$. The null shear $\sigma$ changes a beam's shape at fixed area to first order; it is symmetric and trace-free. The null twist $\omega$ is antisymmetric and measures local rotation of the transverse rays; it vanishes for a hypersurface-orthogonal null geodesic congruence. In $D$ dimensions replace $1/2$ by $1/(D-2)$. These definitions use the trace convention for null expansion, rather than the alternative screen-averaged convention.

Take an affinely parametrized null geodesic congruence, ${\nabla }_{k}k=0$. Choose the auxiliary null vector and a transverse orthonormal screen basis to undergo parallel transport along each ray. For the unprojected tensor $C_{ab}={\nabla }_b{k}_a$, the product rule and the Ricci identity yieldHere the Riemann curvature tensor convention is the one in Question 1; its last-index antisymmetry converts the covector commutator to the displayed curvature term. Because $C_{ab}{k}^a=C_{ab}{k}^b=0$, inserting the screen-space projector in the product term adds only terms killed by these contractions. Parallel propagation makes differentiation commute with the screen projection. Therefore, in screen indices $A,B$,This is the optical matrix evolution equation. Its trace has curvature contraction $q^{ab}{R}_{acbd}{k}^c{k}^d=R_{cd}{k}^c{k}^d$: the additional terms in $q^{ab}-g^{ab}$ vanish by curvature antisymmetry. Write $B=(\theta /2)q+\sigma +\omega$ using the null expansion, null shear, and null twist. Symmetric--antisymmetric cross terms have zero trace, $\sigma$ is trace-free, andThe minus sign in the last trace comes from ${\omega }_{ab}=-{\omega }_{ba}$ and the positive screen metric. Taking the trace therefore derives the Null Raychaudhuri equation:For a nonaffine tangent obeying ${\nabla }_{k}k=\kappa k$, the same product-rule derivation adds $+\kappa \theta$ on the right; the projected derivative of $\kappa$ multiplies $k_a$ and vanishes. In $D$ dimensions the quadratic null expansion term is $-{\theta }^2/(D-2)$.

The screen-space projector annihilates both null vectors: $P^a{}_b{U}^b=P^a{}_b{N}^b=0$, and satisfies $P^2=P$. Its image is the two-dimensional spacelike screen orthogonal to $U,N$. The screen metric isIt is positive definite on that screen. Projecting both indices of $B_{ab}={\nabla }_b{U}_a$ gives the optical tensor ${\hat{B}}_{ab}=P_a{}^c{P}_b{}^d{B}_{cd}$.

Define its three parts byThus ${\hat{B}}_{ab}=\frac{1}{2}\theta q_{ab}+{\sigma }_{ab}+{\omega }_{ab}$. The null expansion $\theta$ is its trace and equals $d\log \mathcal{A}/d\lambda$ for the infinitesimal beam area $\mathcal{A}$. The null shear $\sigma$ is symmetric and trace-free, measuring shape change at fixed first-order area. The null twist $\omega$, also called rotation, is antisymmetric and measures failure of screen directions to remain hypersurface-orthogonal. In $D$ dimensions replace $1/2$ by $1/(D-2)$. Here and below expansion means the trace, rather than its average over the screen dimensions.

For each of the two normal directions, launch a null geodesic congruence orthogonally from the sphere with initial tangent $U=X$ or $U=Y$. For $U=X$ take initial auxiliary $N=Y$; for $U=Y$ take initial $N=X$, and then parallel transport $N$ as in part (a). On the sphere the resulting screen-space projector projects onto its angular tangent space, whose metric is $q_{AB}=r^2{\gamma }_{AB}$.

For any radial normal $K$, the spherical null expansion in ingoing coordinates follows directly from area variation:Equivalently $\mathcal{A}\propto r^2$ gives $K(\log \mathcal{A})=2K(r)/r$. ConsequentlySpherical symmetry also makes the null shear and null twist zero for these radial congruences.

There is a small parametrization distinction. The smooth field $X=-{\partial }_r$ is affine, since ${\Gamma }^a{}_{rr}=0$. The natural smooth field $Y={\partial }_v+(f/2){\partial }_r$ obeys ${\nabla }_{Y}Y=(f^{\prime }/2)Y$. To meet part (a)'s affine convention, use an affine rescaling of a null normal along the outgoing generators, with scaling equal to one on $S$. Its angular derivatives on $S$ then leave the projected derivative unchanged. The displayed null expansions are therefore precisely those for the affinely launched congruences, even though that convenient global expression for $Y$ is nonaffine away from its initial sphere.